Let $\boldsymbol{X} = (X_1,X_2)^{\rm T}\sim \mathcal{N}_2(\boldsymbol{\mu}, \mathrm{\Sigma})$, where \begin{eqnarray*} \boldsymbol{\mu} = (\mu_1, \mu_2)^{\rm T}& = &(\sqrt{\xi_1\xi_2/(\xi_1+\xi_2)}, 0)^{\rm T}\\ \mathrm{\Sigma} & = &\begin{pmatrix} 1 & \rho\\ \rho & 1\end{pmatrix}\\ \rho & = & \sqrt{\xi_1\xi_3/(\xi_1+\xi_2)(\xi_2+\xi_3)}. \end{eqnarray*} It is given that $\xi_1\leq\xi_2\leq\xi_3$, $\xi_i\geq 0$ and $\sum_{i=1}^3\xi_i = 1$. I have a function \begin{equation*} \pi(\boldsymbol{\mu};\boldsymbol{\xi}) = 1\mathbb{P}(\boldsymbol{X}\leq \boldsymbol{t}) = 1\mathbb{P}(X_1\leq t \cap X_2\leq t), \end{equation*} where $t>0$. Under the constraints $\xi_1\leq\xi_2\leq\xi_3$, $\xi_i\geq 0$ and $\sum_{i=1}^3\xi_i = 1$, numerically we are getting the maximum of $\pi(\boldsymbol{\mu};\boldsymbol{\xi})$ where the nonzero mean $\mu_1$ is maximized. This occurs when $\xi_1 = \xi_2 = \xi_3 = 1/3$. Can we use some kind of stochastic ordering arguments to prove the result theoretically?

$\begingroup$ What is the $t$ here? Also, $\mu_1$ is maximized, not when $\xi_1=\xi_2=1/3$, but when $\xi_1=\xi_2=1/2$. $\endgroup$– Iosif PinelisCommented Sep 27, 2021 at 12:35

$\begingroup$ @ Iosif $t>0$. Please look at the restriction $\xi_1\leq\xi_2\leq\xi_3$. $\endgroup$– Satya PrakashCommented Sep 27, 2021 at 14:44
1 Answer
$\newcommand{\der}{\mathrm{der}}\newcommand{\tdert}{\mathrm{dert}}\newcommand{\erf}{\operatorname{erf}}\newcommand{\eqs}{\overset{\text{sign}}=}\newcommand{\tder}{\widetilde\der}$The answer is no.
Indeed, let $x:=\xi_1$ and $y:=\xi_2$, so that $\xi_3=1xy$, $0\le x\le y\le1xy$, whence $x\in[0,1/3]$.
Consider further the case $y=x\in(0,1/3]$, so that $$\mu_1=M(x):=\sqrt{x/2},$$ $$\rho=R(x),\quad R(x):=\sqrt{\frac{12x}{2(1x)}}.$$ Let also $$p_t(x):=1P(X_1\le t,X_2\le t).$$ Your desired result would imply that \begin{equation*} p_t(x)\le p_t(1/3) \tag{0} \end{equation*} for all $x\in(0,1/3]$ and all real $t>0$.
Note that \begin{equation*} p_t(x)=1P(X_1\le t,X_2\le t)=P(M(x),R(x)), \tag{1} \end{equation*} where \begin{equation*} P(m,r):=1\int_{\infty}^t du \int_{\infty}^t dv\, f_{m,r}(u,v) \end{equation*} and $f_{m,r}$ is the density function of the bivariate normal distribution with means $m,0$, variances $1,1$, and correlation $r$.
The key is Plackett's observation (formula (3)) that \begin{equation*} D_r f_{m,r}(u,v)=D_v D_u f_{m,r}(u,v), \end{equation*} where $D_w$ denote the partial derivative with respect to a variable $w$. It follows that \begin{equation*} \begin{aligned} D_r P(m,r)&:=\int_{\infty}^t du \int_{\infty}^t dv \, D_r f_{m,r}(u,v) \\ &=\int_{\infty}^t du \int_{\infty}^t dv \, D_v D_u f_{m,r}(u,v) \\ &=f_{m,r}(t,t). \end{aligned} \tag{2} \end{equation*} Next, \begin{equation*} \begin{aligned} P(m,r)&=1\int_{\infty}^t du \int_{\infty}^t dv\, f_{0,r}(um,v) \\ &=1\int_{\infty}^{tm} dw \int_{\infty}^t dv\, f_{0,r}(w,v) \end{aligned} \end{equation*} and hence \begin{equation*} \begin{aligned} D_m P(m,r)=\int_{\infty}^t dv\, f_{0,r}(tm,v)=\int_{\infty}^t dv\, f_{m,r}(t,v); \end{aligned} \tag{3} \end{equation*} the latter integral can be easily expressed in terms of the error function $\erf$ and elementary functions.
By (1) and a chain rule of differentiation, \begin{equation*} D_x p_t(x)=D_m P(M(x),R(x))M'(x)D_r P(M(x),R(x))R'(x). \end{equation*}
In particular, \begin{equation} D_x p_{1/10}(x)\big_{x=1/3}=\erf\left(\frac{1}{60} \left(3 \sqrt{6}10\right)\right)+1\frac{3 }{\sqrt{\pi }}\,e^{(30 \sqrt{6}77)/1800} =0.739\ldots<0. \end{equation} So, inequality (0) fails to hold for $t=1/10$ and all $x$ in a left neighborhood of $1/3$. $\quad\Box$

$\begingroup$ @Pinelis Thank very much for the detailed reply. I still have few doubts. How is the function $P_{t}(m,r)$ increasing in $m$ not very clear. Moreover, correlation in the question is negative hence the result will be reverse. Also, Slepian inequality applies to random variables with mean 0. Please through some light in these issues. $\endgroup$ Commented Sep 27, 2021 at 18:05

$\begingroup$ @SatyaPrakash : (i) when $m$ is increased by, say, $h$ (keeping $r$ and the variances the same), this amounts to replacing $X_1$ by $X_1+h$. So, $P_t(m,r)$ increases in $m$. (ii) If $r$ is increased, then $r$ is decreased, so that, by Slepian, $P(X_1\le t,X_2\le t)$ is decreased, so that $1P(X_1\le t,X_2\le t)$ is increased. (iii) When varying $r$ only, you apply Slepian's lemma to the centered random variables $X_iEX_i$, which are of mean $0$. $\endgroup$ Commented Sep 27, 2021 at 18:45

$\begingroup$ Previous comment continued: Alternatively, in Slepian's lemma you can replace the pure and mixed second moments by the variances and covariances, respectively, and then the meanzero assumption is not needed there. $\endgroup$ Commented Sep 27, 2021 at 18:46

$\begingroup$ @SatyaPrakash : Actually, the desired result is false in general. $\endgroup$ Commented Sep 29, 2021 at 5:01

$\begingroup$ @Pinelis Actually $t$ is the critical value which is always greater than 1. Can we have some condition over $t$ for which the result holds true? $\endgroup$ Commented Sep 29, 2021 at 6:27